Question: Is ${202824}$ divisible by $9$ ?
Explanation: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {202824}= &&{2}\cdot100000+ \\&&{0}\cdot10000+ \\&&{2}\cdot1000+ \\&&{8}\cdot100+ \\&&{2}\cdot10+ \\&&{4}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {202824}= &&{2}(99999+1)+ \\&&{0}(9999+1)+ \\&&{2}(999+1)+ \\&&{8}(99+1)+ \\&&{2}(9+1)+ \\&&{4} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {202824}= &&\gray{2\cdot99999}+ \\&&\gray{0\cdot9999}+ \\&&\gray{2\cdot999}+ \\&&\gray{8\cdot99}+ \\&&\gray{2\cdot9}+ \\&& {2}+{0}+{2}+{8}+{2}+{4} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${202824}$ is divisible by $9$ if ${ 2}+{0}+{2}+{8}+{2}+{4}$ is divisible by $9$ Add the digits of ${202824}$ $ {2}+{0}+{2}+{8}+{2}+{4} = {18} $ If ${18}$ is divisible by $9$ , then ${202824}$ must also be divisible by $9$ ${18}$ is divisible by $9$, therefore ${202824}$ must also be divisible by $9$.